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Consider a non-associative commutative unital algebra of finite dimension where the product is defined by a Cayley table such that elements are generated with real number coefficients $(a_0, \dots, a_n) $ for a basis $\{1, i_1, \dots, i_n \}$ and we have a basis so that $ i_k^2 \in \{ -1, +1 \}$.

This implies the answer can be given in the form of a Cayley table for multiplying the finite dimensional basis $\{1, i_1, \dots, i_n \}$.

So our algebra is of the form $ a_0 + a_1 i_1 + a_2 i_2 + ...$, addition is as usual and products are defined by a cayley table relating the basis elements $\{1, i_1, \dots, i_n \}$ and we have a basis so that $ i_k^2 \in \{ -1, +1 \}$.

Also the product of two base elements $i_a$ and $i_b$ is never $0$.

Assume the algebra is also power-associative.

One important extra condition :

Let $T$ be such an Algebra.

Let $x$ be an element of $T$. Then $y^2 = x$ always has a solution $y$ within $T$.

So basically the square root is defined in the algebra.

The final condition is that this type of algebra $A$ we seek, is not isomorphic to a tensor product of the complex numbers with another algebra $B$(taken over the reals).

What are simple or small examples of such "hypercomplex" algebra’s ? What is the smallest dimension it can have ?

I know there are methods* to test potential candidates ( dimensions or proposed solutions) but they are cumbersome and give no “ deeper insights “. (*they usually involve things like Jacobi matrix or hessian )

I can’t find examples in literature. So I’m not even sure they exist but I hope they do.

Many thanks.

mick
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  • Does the "not a tensor product with $\mathbb C$ as one factor" condition disquality the octonions? (I don't immediately think it does -- elements of the $\mathbb C$ factor would commute with everything, whereas only the reals have this property in $\mathbb O$. But I might be overlooking something). – Troposphere Mar 29 '21 at 19:53
  • @Troposphere the octonions are NOT commutative – mick Mar 29 '21 at 19:58
  • Ah, sorry, I missed that you require commutativity. – Troposphere Mar 29 '21 at 19:58
  • This is just a rearrangement of Troposphere's answer. What if you take $\mathbb{O}$ and define a new product by $x\star y=\frac{xy+yx}{2}$? It is commutative now. – 2ndYearFreshman Mar 31 '21 at 00:06
  • @2ndYearFreshman ok that is commutative. But how about all the other requirements ?? – mick Mar 31 '21 at 21:12

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